1. Free Body Analysis & Equations of Motion

We analyze the system in two parts: the hanging masses and the rolling wheel.

Part A: Hanging Masses (A + B)
Total mass = $m + 3m = 4m$. Let acceleration be $a$ downwards.
Equation of motion:

$$ 4mg – T = 4ma \quad \dots(1) $$

Where $T$ is the tension in the main string.

Part B: Rolling Wheel
Mass $m$, Radius $R$. The wheel rolls without slipping ($a = \alpha R$).
Forces acting horizontally: Tension $T$ (right), Spring force $kx$ (left), Friction $f$ (left).
Translational equation:

$$ T – kx – f = ma \quad \dots(2) $$

Rotational equation (Torque about center):

$$ \tau = f R = I \alpha $$

Since mass is uniformly distributed on the rim, $I = mR^2$.

$$ f R = (mR^2) \left( \frac{a}{R} \right) \implies f = ma $$

Substituting $f$ into equation (2):

$$ T – kx – ma = ma \implies T – kx = 2ma \quad \dots(3) $$

2. System Dynamics (SHM)

Substitute $T$ from eq (1) into eq (3):

$$ (4mg – 4ma) – kx = 2ma $$ $$ 4mg – kx = 6ma $$ $$ a = \frac{4mg – kx}{6m} $$

This is the equation of Simple Harmonic Motion (SHM) of the form $a = \omega^2 (x_0 – x)$.
At equilibrium ($a=0$), the extension is $x_0 = \frac{4mg}{k}$.
The motion starts from rest at $x=0$ (unstretched spring). Thus, $x=0$ is an extreme position, and the amplitude of oscillation is $A = x_0$.

The acceleration ranges between extreme values:

• At $x = 0$ (Start): $a_{start} = \frac{4mg}{6m} = \frac{2g}{3}$ (Downwards).
• At $x = 2x_0$ (Max extension): $a_{min} = \frac{4mg – k(8mg/k)}{6m} = \frac{-4mg}{6m} = -\frac{2g}{3}$ (Upwards).

3. Maximum Tension Calculation

We need to find the maximum tension in the thread connecting loads A and B (let’s call it $T_{AB}$).
Consider the Free Body Diagram of block B (mass $3m$):

$$ 3mg – T_{AB} = 3m a $$ $$ T_{AB} = 3m(g – a) $$

To maximize $T_{AB}$, the term $(g – a)$ must be maximum. This occurs when acceleration $a$ is minimum (i.e., maximum negative value, acting upwards).

$$ a_{min} = -\frac{2g}{3} $$

Substituting this value:

$$ T_{AB(max)} = 3m \left( g – \left( -\frac{2g}{3} \right) \right) $$ $$ T_{AB(max)} = 3m \left( g + \frac{2g}{3} \right) = 3m \left( \frac{5g}{3} \right) $$ $$ T_{AB(max)} = 5mg $$

Final Answer

The maximum tension developed in the thread connecting the loads is:

$$ 5mg $$